A square loop of side 10 cm and resistance 0.7 Ω is placed vertically… | JEE Main 2024 (29 Jan, Shift 1)

Q38JEE Main 2024 · 29 JAN · SHIFT 1Medium

Magnetic Flux & Faraday's Law

A square loop of side $10\text{ cm}$ and resistance $0.7\,\Omega$ is placed vertically in east-west plane. A uniform magnetic field of $0.20\text{ T}$ is set up across the plane in north east direction. The magnetic field is decreased to zero in $1\text{ s}$ at a steady rate. Then, magnitude of induced emf is $\sqrt{x} \times 10^{-3}\text{ V}$ . The value of $x$ is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Loop in the east-west vertical plane, so its normal points north-south.
Field along north-east makes $45^\circ$ with the normal: $\phi = BA\cos 45^\circ = 0.2 \times 0.01 \times \frac{1}{\sqrt{2}} = \sqrt{2} \times 10^{-3}$ Wb.
Flux falls to zero steadily in 1 s: $e = \frac{\sqrt{2} \times 10^{-3}}{1} = \sqrt{2} \times 10^{-3}$ V.
So $x = 2$ .

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